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Let $A,B$ be abelian groups. Let $A[2]=\{a\in A\mid 2a=0\}$. Let $f:A \to B$ be a surjective homomorphism. Let $f':A[2]\to B[2]$ be the induced group homomorphism. When is $f'$ surjective ?

In the case $A=A[2]$ holds, it is clearly surjective. But if $A\neq A[2]$, what kind of condition ensures $f'$ is surjective ?

Background: Let $K$ be a quadratic number field. There is surjective map from $Cl(K) \to Cl(K,S)$ where $Cl(K,S)$ is $S$-ideal class group (Relation between $S$-ideal class group and usual ideal class group). I want to know when $Cl(K)[2]\to Cl(K,S)[2]$ is surjective. Especially, I want to prove $Cl(K)[2]=0 \implies Cl(K,S)[2]=0$.

user26857
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Pont
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  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – Shaun Jul 16 '23 at 18:47
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    @citadel. $2a$ means $a+a$ in $(A,＋)$. – Pont Jul 16 '23 at 18:55

1 Answers1

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$ Cl(K)[2]=1$ implies $Cl(K)$ is odd abelian group. There is surjection $Cl(K) \to Cl(K,S)$, then order of $Cl(K,S)$ is odd. Thus $Cl(K,S)[2]=1$.

Pont
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